What are the solutions to the equation x=(6-x)^2 ?

The solutions to the equation x=(6-x)^2 are x=9,x=4

Find the zeros of x=(6-x)^2

The zeros of x=(6-x)^2 are x=9,x=4

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x = (6 - x)^{2}

x = 9, x = 4

Solution steps

x = (6 - x)^{2}

Expand

(6 - x)^{2} : 36 - 12 x + x^{2}

x = 36 - 12 x + x^{2}

Switch sides

36 - 12 x + x^{2} = x

Move

x

to the left side

x^{2} - 13 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 36 = 5

x_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 1}

x = \frac{- ( - 13 ) + 5}{2 \cdot 1} : 9

x = \frac{- ( - 13 ) - 5}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 9, x = 4

- x^{2} - 5 x + 14 = 0

4 n^{2} + 3 n = 3

- 2 x^{2} + 3 x - 7 = 0

8 x^{2} - 6 x - 1 = 0

100 x^{2} + 3 = 7

What are the solutions to the equation x=(6-x)^2 ?
The solutions to the equation x=(6-x)^2 are x=9,x=4
Find the zeros of x=(6-x)^2
The zeros of x=(6-x)^2 are x=9,x=4